Vtyjkyuk

The differential equation is as follows

$$begincases
y'' + y = 8e^-2t sin t \
y(0)=0 \y'(0)=0
endcases$$

How do I solve it by Laplace transformation?

In my solution I've taken the Laplace transform of both sides and got to the solution of the transformed equation $$mathcal Ly = frac8[(s-2)^2+1](s^2+1)$$ but I'm stuck, I don't know how to take the inverse Laplace transform to get the solution to the DE.

There are two methods that can be used. The first being the convolution and the second being the longer, more standard, version.
Let the Laplace transform be given by
$$mathcalLy(t) = f(s) = int_0^infty e^-s t , y(t) , dt.$$
Since,
$$mathcalL y'' + y = s^2 , f - s y(0) - y'(0) + f = (s^2 + 1) , f,$$
then by convolution the solution takes the form:
beginalign
y(t) = 8 , int_0^t cos(t-u) , e^-2 u , sin(u) , du.
endalign
This is seen by
beginalign
(s^2 + 1) , f &= mathcalL8 , e^-2t , sin(t) \
f &= frac1s^2 + 1 , mathcalL8 , e^-2t , sin(t) \
&= mathcalLcos(t) cdot mathcalL8 , e^-2t , sin(t)
endalign
and after the inversion, by convolution, the integral form, which can be calculated, is obtained.

The more standard version is:
beginalign
(s^2 + 1) , f &= mathcalL8 , e^-2t , sin(t) = frac8(s+2)^2 + 1 \
f &= frac8(s^2 + 1) , ((s+2)^2 + 1) \
&= frac1-ss^2 + 1 + fracs+3(s+2)^2 + 1 \
&= frac1s^2 + 1 - fracss^2+1 + fracs+2(s+2)^2 + 1 + frac1(s+2)^2 + 1
endalign
and leads to
$$y(t) = sin t - cos t + e^-2 t , (sin t + cos t).$$

Both methods yield the same result after some calculations are applied for the convolution case.

Hint

Apply Laplace transform to both sides of the equation to get $$ s^2 tilde y -s underbracey(0^+) - y'(0^+)_textZero from the\textinitial conditions + tilde y = frac8(s+2)^2+1 \[20 pt] s^2tilde y + tilde y = frac8(s+2)^2+1$$
Where $tilde y = mathcal Ly$. Now you can easily solve for $tilde y$ and do an inverse Laplace transform to find the solution $y$.

By simplifying the solution you'll get $$tilde y = frac8[(s+2)^2+1](s^2+1)$$ and by doing partial fraction decomposition $$tilde y = fracAs+B(s+2)^2+1+fracCs+Ds^2+1 = frac(As+B)[(s+2)^2+1]+(Cs+D)(s^2+1)[(s+2)^2+1](s^2+1)$$ Now just find the values of $A,B,C,D$ and then take the inverse Laplace of the decomposed function.